In a hand of 13 playing cards from a deck of 52 whats the probability of drawing exactly one king.
My approach would be $${4 \choose 1}*{48 \choose 12}/{52 \choose 13}*{2}$$
I divided by 2 because I felt I had ordered the king and the other 12 cards chosen but this is wrong. Can someone please explain in depth why this is wrong so that I don't make the same mistake again.
On
The total number of ways to choose $13$ out of $52$ cards is:
$$\binom{52}{13}$$
The number of ways to choose $13$ out of $52$ cards with exactly $1$ king is:
$$\binom{4}{1}\cdot\binom{52-4}{13-1}$$
Hence the probability of choosing $13$ out of $52$ cards with exactly $1$ king is:
$$\frac{\binom{4}{1}\cdot\binom{52-4}{13-1}}{\binom{52}{13}}\approx43.88\%$$
you have two disjoints subsets: 'kings' and 'everything else'. $\binom{4}{1}\binom{48}{12}$ simply means 'any 1 out of 4' AND 'any 12 out of 48'. There's no order involved. Any form of order would be if, for example, you would need to get 3 kings. Then you would need to divide by $3!$ because the order doesn't matter.
Does this answer your question?